What is the integral of (9x)/(x^3-3x+2) ?

The integral of (9x)/(x^3-3x+2) is 2ln|x-1|-3/(x-1)-2ln|x+2|+C

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integral of (9x)/(x^3-3x+2)

\int \frac{9 x}{x ^{3} - 3 x + 2} d x

2 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 2 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{9 x}{x ^{3} - 3 x + 2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{x}{x ^{3} - 3 x + 2} d x

Take the partial fraction of

\frac{x}{x ^{3} - 3 x + 2} : \frac{2}{9 ( x - 1 )} + \frac{1}{3 ( x - 1 ) ^{2}} - \frac{2}{9 ( x + 2 )}

= 9 \cdot \int \frac{2}{9 ( x - 1 )} + \frac{1}{3 ( x - 1 ) ^{2}} - \frac{2}{9 ( x + 2 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (\int \frac{2}{9 ( x - 1 )} d x + \int \frac{1}{3 ( x - 1 ) ^{2}} d x - \int \frac{2}{9 ( x + 2 )} d x)

\int \frac{2}{9 ( x - 1 )} d x = \frac{2}{9} ln ∣ x - 1 ∣

\int \frac{1}{3 ( x - 1 ) ^{2}} d x = - \frac{1}{3 ( x - 1 )}

\int \frac{2}{9 ( x + 2 )} d x = \frac{2}{9} ln ∣ x + 2 ∣

= 9 (\frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} ln ∣ x + 2 ∣)

Simplify

9 (\frac{2}{9} ln ∣ x - 1 ∣ - \frac{1}{3 ( x - 1 )} - \frac{2}{9} ln ∣ x + 2 ∣) : 2 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 2 ln ∣ x + 2 ∣

= 2 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 2 ln ∣ x + 2 ∣

Add a constant to the solution

= 2 ln ∣ x - 1 ∣ - \frac{3}{x - 1} - 2 ln ∣ x + 2 ∣ + C

What is the integral of (9x)/(x^3-3x+2) ?
The integral of (9x)/(x^3-3x+2) is 2ln|x-1|-3/(x-1)-2ln|x+2|+C